CP2 ELECTROMAGNETISMhttps://users.physics.ox.ac.uk/∼harnew/lectures/

LECTURE 20:

ELECTROMAGNETICWAVES

Neville Harnew1

University of Oxford

HT 2020

1With thanks to Prof Laura Herz1

OUTLINE : 20. ELECTROMAGNETIC WAVES

20.1 Divergence, time derivative, and curl of E and B

20.2 Electromagnetic waves : speed of propagation

20.3 Relationship between E and B

20.4 Electromagnetic wave travelling along the z direction

20.5 Characteristic impedance of free space

20.6 Polarisation

20.7 Energy flow and the Poynting Vector

20.8 Example : Poynting Vector for a long resistive cylinder

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20.1 Divergence, time derivative, and curl of E and B

I The divergence of E : ∇ ·E = ∇ ·E0 exp [i(ωt − k · r)]

=[∂∂ x ,

∂∂ y ,

∂∂ z

]·E0 exp

(i(ωt − kxx − kyy − kzz)

)=[(−i) kxEx + (−i) kyEy + (−i) kzEz

]exp (i(ωt − k · r))

= (−i)k ·E : hence ∇ ≡ −i k

I The time derivative of E : ∂∂ t E = ∂

∂ t E0 exp [i(ωt − k · r)]

= i ωE : hence ∂∂ t ≡ i ω

I The curl of E :

∇×E =

∣∣∣∣∣∣i j k∂∂ x

∂∂ y

∂∂ z

Ex Ey Ez

∣∣∣∣∣∣ =

∂Ez∂ y −

∂Ey∂ z

∂Ex∂ z −

∂Ez∂ x

∂Ey∂ x −

∂Ex∂ y

=

(−i)

ky Ez − kzEykzEx − kxEzkxEy − ky Ex

= (−i)k×E & again ∇ ≡ −i k

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20.2 Electromagnetic waves : speed of propagation

I To get speed of propagation, substituteE = E0 exp (i(ωt − k · r)) into the wave equation∇2 E = ε0 µ0

∂2

∂ t2E

I Use ∇ ≡ −i k → ∇2 ≡ (−i k)2 = −k2

∂∂ t ≡ i ω → ∂2

∂ t2 ≡ (i ω)2 = −ω2

I −k2 E0 exp (i(ωt − k · r)) = −ω2 ε0 µ0 E0 exp (i(ωt − k · r))

→ k2 = ω2 ε0 µ0

I Fields of this form are solutions to the wave equation withvelocity of propagation :c = ω

k = 1√ε0 µ0

= 3× 108 m s−1

i.e. the speed of light→ speed of an EM wave in vacuum

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20.3 Relationship between E and BI Substitute E = E0 exp (i(ωt − k · r)) into Maxwell eqn’s :∇ ·E = −i k ·E = 0∇ ·B = −i k ·B = 0Hence k ·E = 0 and k ·B = 0

I Electric and magnetic fields in vacuum are perpendicularto direction of propogation→ EM waves are transverse

I Substitute into Faraday’s Law : ∇×E = −∂B∂t

−i k×E = −i ωB → B = 1ω k×E

I Substitute into Ampere’s Law : ∇×B = µ0 ε0∂E∂t

−i k×B = i ω µ0 ε0 E → E = −c2

ω k×B

I E, B and k are mutually orthogonalI E and B are in phase and lie in the plane of the wavefront

I Field magnitude ratio : |E|/|B| = c2

ω k = c = 1√µ0 ε0

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20.4 Electromagnetic wave travelling along the z direction

E = E0 sin (ω(t − z/c)) ex

B = B0 sin (ω(t − z/c)) ey6

20.5 Characteristic impedance of free space

I Take the ratio Z = |E||H| where |H| = 1

µ0|B|

I Z has units [V m−1] / [ A m−1 ] = Ohms.

I Z is called the characteristic impedance offree space

Z = µ0|E||B| = µ0 c = µ0√

µ0 ε0=√

µ0ε0

= 376.7 Ω

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Electromagnetic waves : summary

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20.6 PolarisationI Linearly (or plane) polarised wave :

E has one specific orientationI Circularly polarised wave :

Two linear components of Esuperimposed at a right angle andphase shifted by π/2 E00

sin(ωt) +

0E0

sin(ωt + π/2) = E

sin(ωt)cos(ωt)

0

I Elliptically polarised wave :

As above but with unequal amplitudesI Unpolarised :

E superimposed with all orientations(with no fixed phase relationshipsbetween components)

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20.7 Energy flow and the Poynting VectorI Recall : Energy of the electric field Ue =

∫ν

12 ε0 E

2 dνRecall : Energy of the magnetic field Um =

∫ν

12µ0

B2 dνI Total EM energy in volume ν :

U =∫ν

12

(ε0 E ·E+

1µ0

B ·B)

︸ ︷︷ ︸Energy density

dν

I In free space ( J = 0, ρ = 0)∂E

∂t=

1µ0ε0

∇×B︸ ︷︷ ︸Ampere’s Law

;∂B

∂t= −∇×E︸ ︷︷ ︸

Faraday’s Law

I Calculate the rate of change of energy in ν :dUdt =

∫ν

(ε0 E · E+ 1

µ0B · B

)dν

dUdt =

∫ν

(ε0µ0ε0

(E · ∇ ×B)− 1µ0

B · (∇×E))

dνdUdt = − 1

µ0

∫ν ∇ · (E×B)dν

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Energy flow and the Poynting Vector continuedI Energy flow out of volume ν per unit time :

dUdt = − 1

µ0

∫ν ∇ · (E×B)dν

I Apply the divergence theorem :

dUdt = −

∮S (

1µ0

E×B)︸ ︷︷ ︸Poynting Vector, N

· da

dUdt = −

∮S N · da where N = 1

µ0E×B

Poynting vector N is the power per unit area flowingthrough the surface bounded by volume ν . (It also givesthe direction of flow). Units of N : [W m−2]

I For EM waves, the intensity is the time-average of |N|= =< |N| >= 1

µ0E0 B0< cos2(ωt − k · r) >︸ ︷︷ ︸

1/2

= 12µ0 c E2

0

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20.8 Example : Poynting Vector for a long resistive cylinderI Calculate Poynting Vector at the surface of the wire with

applied potential difference V and current IN = 1

µ0E×B

I Electric field along wire axis : E = V/`

Magnetic flux density at wire surface :∮B · d` = B · 2πa = µ0I

(note that this is tangential - alongcircumference)

I N = 1µ0

V`µ0I2πa

(in radial direction pointing inwards -i.e. wire heats up !)

I Hence N = (V I)/2π`a︸ ︷︷ ︸Hence N = (V ))surface area

I Total power dissipated in wire : P =∫

S N · da = V I

as expected from circuit theory.12

Poynting Vector : summary

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